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Classify all Maximal Ideals on this Ring!

  1. Mar 10, 2012 #1
    Classify all maximal ideals on the commutative ring of continuous functions on [0,1] (C[0,1],+,.)

    I am not too confident about my solution. Can someone please look through it?

    Fortunate guess:-
    The set J(y) = {f in C[0,1] / f(y)=0} where y is fixed in [0,1]

    This is an ideal because
    (1) If g,h belong to J(y) then g-h belongs to J(y) => (J(y),+) is a subgroup of (C[0,1],+)
    (2) If g is in J(y) and h is in C[0,1] then g.h is in J(y)

    Also, C[0,1]/J(y) is a field because,
    If J(y)+f is in C[0,1]/J(y) and J(y)+f =/= J(y), then f(y) =/= 0.
    Let g(x)=1/f(y) for all x in C[0,1]
    Then, (J(y)+g).(J(y)+f)=(J(y)+f.g)=(J(y)+1) [because f.g-1 is in J(y)]
    Hence (J(y)+f) is invertible.

    And, C[0,1]/J(y) is a field => J(y) is a maximal ideal.

    To show there are no other maximal ideals:-

    Let I be an ideal of C[0,1] such that I is not {0} and I is not a subset of any J(y)

    That implies, for each z in [0,1], there exists f in I such that f(z) =/= 0.
    multiplying this f by an appropriate constant function AND an appropriate continuous piecewise linear function will guarantee, for each z, the existence of a function g in I such that g(z)>1 AND g>0 on [0,1].

    for z in [0,1], the corresponding function g and e=1/2, there exists a number d(z)>0 such that in B(z,d(z)) g is always greater than 1/2.

    B(z,d(z)) is an open cover for [0,1] so it must admit a finite subcover. This will give me a finite number of functions g1, g2,......, gk in I such that their sum is a function in I that never becomes zero. Multiplying this function by its reciprocal will imply that the identically 1 function is in I. Which implies I=C[0,1]

    Hence the J(y)'s are the only maximal ideals in C[0,1] (?)

    Thank you for reading the whole thing.
     
  2. jcsd
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